Pollard's kangaroo ECDLP solver

[Alpaste]

Yes you are right, but one point i don't get it.
why would i scan 2^256 range? isn't it  scanning only 2^160 using for example BSGS to find any specific publickey with its private key? (since all publickeys and all private keys are in the range of 2^160.)

[arulbero]

If we talking about bruteforce like bitCrack, you have to check 2^97 keys in average to find specific address (not the specific public key)

No, you have to check 2^(256-96) = 2^160 keys in average to find the private key to a specific address.

Yes you are right, but one point i don't get it.
why would i scan 2^256 range? isn't it  scanning only 2^160 using for example BSGS to find any specific publickey with its private key? (since all publickeys and all private keys are in the range of 2^160.)

More or less. You don't know for sure that all addresses have a private key in range [1, ..., 2^160]

[_Counselor]

No, you have to check 2^(256-96) = 2^160 keys in average to find the private key to a specific address.

yeah, you're right here, I wrote it wrong.

[Alpaste]

arulbero, i'm wondering, how does these 2^96 collisions happen? does the 2^96 addresses each address share all the same public key or how does 2^96 addresses can generate the same address?

[arulbero]

arulbero, i'm wondering, how does these 2^96 collisions happen? does the 2^96 addresses each address share all the same public key or how does 2^96 addresses can generate the same address?

The same address is related to 2^96 different public keys (and 2^96 different private keys)

An address is only the hash of a public key; different public keys (256 bit) have the same hash (160bit).

[Alpaste]

That's insane! Thank you for clarifying!
may i ask but, how is it possible to an same address to relate to another completely different public keys and private keys? how do they match to same address, if they're completely different?
Thanks again.

[arulbero]

That's insane! Thank you for clarifying!
may i ask but, how is it possible to an same address to relate to another completely different public keys and private keys? how do they match to same address, if they're completely different?
Thanks again.

A: the set of 2^256 public keys
B: the set of 2^160 addresses (hashes of the public keys)

the hash function maps the set A into the set B

hash :  A -> B

when the # of keys > # of hashes, collisions are inevitable


source: https://en.wikipedia.org/wiki/Hash_function

[akaki]

Hi,

One question regarding the performance of the algorithm.

It's stated that for puzzle #120, the :

Expected time: ~2 months on 256 Tesla V100.

This is really mind boggling  . But does someone know how long it took  (or a time approximation) to solve puzzle #115 using 4 Tesla V100 ?

To me it took (60/2^5)*(256/4)= 4 months, which is surely not reasonnable.

[mausuv]

https://github.com/ragestack/EC-Point-Operations/blob/master/EC_Math.py
https://bitcointalk.org/index.php?topic=5244940.msg58488513#msg58488513

Code:

Point1 = (x1,y1)
Point2 = (x2,y2)

Point3 = sub(Point1,Point2) 
Point4 = add(Point1,Point2)
Point5 = div(Point1,2) # remove Scalar option, i need two point use div Point5 = div(Point1,Point2)
Point6 = mul(Point1,2) # remove Scalar option, i need two point use mul Point6 = mul(Point1,Point2)

i need two point use div Point5 = div(Point1,Point2) #remove Scalar option
i need two point use mul Point6 = mul(Point1,Point2) # remove Scalar option
eny method Point1,Point2 use div and mul get therd point #mybadenglish
read my top 2 link modify please..

its posible remove scalar

Code:

#modify please
def ECdiv(Qx,Qy,Scalar): # EC point division remove Scalar option
    A = (N-1)/Scalar
    Px,Py = ECmul(Qx,Qy,A)
    Py = P-Py
    return Px,Py
Dx,Dy = ECdiv(Cx,Cy,2) 
print (Dx,Dy)

[MrFreeDragon]

https://github.com/ragestack/EC-Point-Operations/blob/master/EC_Math.py
https://bitcointalk.org/index.php?topic=5244940.msg58488513#msg58488513

-snip-
i need two point use div Point5 = div(Point1,Point2) #remove Scalar option
i need two point use mul Point6 = mul(Point1,Point2) # remove Scalar option
eny method Point1,Point2 use div and mul get therd point #mybadenglish
read my top 2 link modify please..
-snip-

Use the whole code from the post https://bitcointalk.org/index.php?topic=5244940.msg58488513#msg58488513

In order to add the division function, you can just use the same multiplication function with the inverse of the scalar.
However you can not divide one point by another point, You can divide only point by a scalar.

[jacky19790729]

Hi,
One question regarding the performance of the algorithm.
It's stated that for puzzle #120, the :

Expected time: ~2 months on 256 Tesla V100.

This is really mind boggling  . But does someone know how long it took  (or a time approximation) to solve puzzle #115 using 4 Tesla V100 ?
To me it took (60/2^5)*(256/4)= 4 months, which is surely not reasonnable.

#110 about  2  days on 256 Tesla V100
#115 about 11 days on 256 Tesla V100

4 Tesla V100 ~~ should be 256/4=64
It takes 64 times ~~~
about 1 day (very lucky) ~ 704 days ( maybe it will take longer )

i need two point use div Point5 = div(Point1,Point2) #remove Scalar option
i need two point use mul Point6 = mul(Point1,Point2) # remove Scalar option
eny method Point1,Point2 use div and mul get therd point #mybadenglish
read my top 2 link modify please..
its posible remove scalar

Code:

python 3


Gx = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
Gy = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
PG = Point(Gx, Gy)
P1 = Point(0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798,0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8)
P2 = Point(0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5,0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a)
P3 = Point(0xf9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9,0x388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672)
P4 = Point(0xe493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13,0x51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922)


# Puzzle 63  $$$  1NpYjtLira16LfGbGwZJ5JbDPh3ai9bjf4
Puzzle_63  = Point(0x65ec2994b8cc0a20d40dd69edfe55ca32a54bcbbaa6b0ddcff36049301a54579, 0x5a1b76ab01e9edd0de24157ceff77bcb0f615560b250b365a5d435873eaa4625 )     

# Puzzle 120 $$$  17s2b9ksz5y7abUm92cHwG8jEPCzK3dLnT
Puzzle_120 = Point(0xceb6cbbcdbdf5ef7150682150f4ce2c6f4807b349827dcdbdd1f2efa885a2630, 0x2b195386bea3f5f002dc033b92cfc2c9e71b586302b09cfe535e1ff290b1b5ac )

# Puzzle 115  $$$ 
Puzzle_115 = mulk( 0x60F4D11574F5DEEE49961D9609AC6, P1 )
print ("[#115]  %064x  %064x" % (Puzzle_115.x,Puzzle_115.y) )

P_add = add( P1 , P2 )
print ("[add ]  %064x  %064x" % (P_add.x,P_add.y) )

P_sub = sub( P3 , P1 )
print ("[sub ]  %064x  %064x" % (P_sub.x,P_sub.y) )

P_mul2 = mul2( P1 )
print ("[mul2]  %064x  %064x" % (P_mul2.x,P_mul2.y) )

P_mulk = mulk( 0x3, P1 )
print ("[mulk]  %064x  %064x" % (P_mulk.x,P_mulk.y) )

# 0x4 / 0x2 = 0x2
P_div = div( P4 , 0x2 )
print ("[div ]  %064x  %064x" % (P_div.x,P_div.y) )

# 0x4 / 0x4 = 0x1
P_div = div( P4 , 0x4 )
print ("[div ]  %064x  %064x" % (P_div.x,P_div.y) )

# 0x7CCE5EFDACCF6808 / 0x3E672F7ED667B404 = 0x2 
P_div = div( Puzzle_63 , 0x3E672F7ED667B404 )
print ("[div ]  %064x  %064x" % (P_div.x,P_div.y) )

===========output============
[#115]  48d313b0398d4923cdca73b8cfa6532b91b96703902fc8b32fd438a3b7cd7f55  66f0742c24f5ff80c799d691d756ad8e5aa7f6d8f986abd9eeef45514637c0d4
[add ]  f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9  388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
[sub ]  c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5  1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
[mul2]  c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5  1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
[mulk]  f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9  388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
[div ]  c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5  1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
[div ]  79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798  483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
[div ]  c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5  1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a

[darangus]

Hi all, i've started up a dedicated kangaroo box, and i was trying all different combinations, i was just wondering what would be a good random combination to just leave running for a year, i'm looking to strike it lucky, not looking to solve specific puzzles.

Throw some ideas at me.

[fxsniper]

Can possible do calculate kangaroo by do manual ?
puzzle120 I would like to try my range by do manual made kangaroo

tame and wild is public key (point) and do multiply to number right?
I will try do python script  generate tame and wide each one a million line of set
and compare it both by manual too

[batareyka]

Just silly question  

is it possible to know this public key is from this range? like 110 or 115?

is there any way to identify?

No, otherwise you would be able to find the upper bits of every private key in existence.

Hi. Can you explain how you can learn the upper bits by knowing the range?

Hi. I once asked the question of whether it is possible to calculate mathematically in which range the key is.
To which I received a response from NotATether.
If you know the range you can calculate the upper bits of the key.
The question logically asks for answers.
If we know, and we know in what range the key lies (like puzzle 120).
Then why not calculate the upper bits?
What is the calculation algorithm?
Thanks.

[Alpaste]

I think it's impossible to calculate the upper bits of the key, even if you know that this private key lies in the 2^120 bits-range.
The only way to get the key, is to use kangaroo, or BSGS, but the chances that you find the key is extremely low.
Cheers!

[batareyka]

Alpaste
I also hold this opinion. But I doubted, re-reading the forum posts because NotATether claimed that it is possible to calculate ..
NotATether can dispel the myth?

[Alpaste]

Alpaste
I also hold this opinion. But I doubted, re-reading the forum posts because NotATether claimed that it is possible to calculate ..
NotATether can dispel the myth?

This claim might be not true at all. If it's possible then, why he himself "NotATether" didn't calculate the private key for the puzzle transactions and withdraw its funds?

[batareyka]

Let's wait, maybe NotATether will answer and dispel the myth.

[Feron]

Hi all this code guess random public key problem this code is slow play with it and try to improve and speed it up
code:
import collections
import random

EllipticCurve = collections.namedtuple('EllipticCurve', 'name p a b g n h')
# here we define our elliptic curve
curve = EllipticCurve(
    'secp256k1', # curve type
    # Field characteristic.
    p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f,
    # Curve coefficients.
    a=0,
    b=7,
    # Define the base point.
    g=(0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798, 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8),
    # Subgroup order.
    n=0x000000000000000000000000000000000000000000000000000000000000ffff,######important###### this manipulate public key maximum guess range up and down
    # Subgroup cofactor.
    h=1,)
# Modular arithmetics application
def inverse_mod(k, p):
    # Returns the inverse of k mod p
    # returns x such that (x * k) % p == 1
    # k <> 0 and p = prime
    if k == 0:
        raise ZeroDivisionError('division by zero') # zero division error!
    if k < 0:
        # k ** -1 = p - (-k) ** -1  (mod p)
        return p - inverse_mod(-k, p) # return the inverse
    # Extended Euclidean algorithm
    s, old_s = 0, 1
    t, old_t = 1, 0
    r, old_r = p, k
    while r != 0:
        quotient = old_r // r
        old_r, r = r, old_r - quotient * r
        old_s, s = s, old_s - quotient * s
        old_t, t = t, old_t - quotient * t
    gcd, x, y = old_r, old_s, old_t
    assert gcd == 1
    assert (k * x) % p == 1
    return x % p
# Functions applied on curve points
def is_on_curve(point):
    # Returns True if the given point lies on the elliptic curve
    if point is None:
        # None represents the point at infinity.
        return True
    x, y = point
    return (y * y - x * x * x - curve.a * x - curve.b) % curve.p == 0
def point_neg(point):
    # Returns -point
    assert is_on_curve(point)
    if point is None:
        # -0 = 0
        return None
    x, y = point
    result = (x, -y % curve.p)
    assert is_on_curve(result)
    return result
def point_add(point1, point2):
    # Returns the result of (P1 + P2) according to the group law
    assert is_on_curve(point1)
    assert is_on_curve(point2)
    if point1 is None:
        # 0 + P2 = P2
        return point2
    if point2 is None:
        # P1 + 0 = P1
        return point1
    x1, y1 = point1
    x2, y2 = point2
    if x1 == x2 and y1 != y2:
        # P1 + (-P2) = 0
        return None
    if x1 == x2:
        # if (P1 == P2)
        m = (3 * x1 * x1 + curve.a) * inverse_mod(2 * y1, curve.p)
    else:
        # if (P1 != P2).
        m = (y1 - y2) * inverse_mod(x1 - x2, curve.p)
    x3 = m * m - x1 - x2
    y3 = y1 + m * (x3 - x1)
    result = (x3 % curve.p,
              -y3 % curve.p)
    assert is_on_curve(result)
    return result
def scalar_mult(k, point):
    # Returns (k * P) computed using the double-and-add algorithm
    assert is_on_curve(point)
    if k % curve.n == 0 or point is None:
        return None
    if k < 0:
        # k * P = -k * (-P)
        return scalar_mult(-k, point_neg(point))
    result = None
    addend = point
    while k:
        if k & 1:
            # Add
            result = point_add(result, addend)
        # Double
        addend = point_add(addend, addend)
        k >>= 1
    assert is_on_curve(result)
    return result
# Keypair generation and ECDHE
def make_keypair():
    # Generate random private-public key pair
    private_key = random.randrange(10000, curve.n)######important###### this equivalent start public key range manipulate this you up and down start public key range momental start is 10000 decimal
    public_key = scalar_mult(private_key, curve.g)
    return private_key, public_key
for xxx in range(1000000):
 aaaa_private_key, aaaa_public_key = make_keypair()
 bbbb_private_key, bbbb_public_key = make_keypair()
 if (str("02{:x}".format(*aaaa_public_key))).endswith("7a"):
  print("aaaa' private key:",hex(aaaa_private_key),"02{:x}".format(*aaaa_public_key),xxx)
 if (str("02{:x}".format(*aaaa_public_key))) == "029d8c5d35231d75eb87fd2c5f05f65281ed9573dc41853288c62ee94eb2590b7a": #any public key use this line# i use test key
  f=open("von.txt","a")
  f.write(str(aaaa_private_key)+"<-decimal key-"+(str("02{:x}".format(*aaaa_public_key)))+"\n")
  f.close()
 if (str("02{:x}".format(*bbbb_public_key))).endswith("7a"):
  print("bbbb' private key:",hex(bbbb_private_key),"02{:x}".format(*bbbb_public_key),xxx)
 if (str("02{:x}".format(*bbbb_public_key))) == "029d8c5d35231d75eb87fd2c5f05f65281ed9573dc41853288c62ee94eb2590b7a": #any public key use this line# i use test key
  f=open("von.txt","a")
  f.write(str(bbbb_private_key)+"<-decimal key-"+(str("02{:x}".format(*bbbb_public_key)))+"\n")
  f.close()
  # aaaa and bbbb now exchange their public keys and verify the shared secret
  s1 = scalar_mult(aaaa_private_key, bbbb_public_key)
  s2 = scalar_mult(bbbb_private_key, aaaa_public_key)
  assert s1 == s2

[NotATether]

Hi all this code guess random public key problem this code is slow play with it and try to improve and speed it up
...

I know this probably sounds very trivial but a good place to start optimization is to convert that Python code to C/C++ using Bitcoin's libsecp256k1 and libbitcoin (I think Bitcoin Core exports its address functions to libbitcoin).